DC Pandey Solutions: Wave Motion

Introductory Exercise 14.1

Q.1. Prove that the equation y = a sin wt does not satisfy the wave equation and hence it does not represent a wave.
Sol.

The one-dimensional wave equation is
∂²y/∂x² = (1/v²) ∂²y/∂t², where v is the wave speed.
For y = a sin ωt, the spatial derivative is
∂²y/∂x² = 0
but the time derivative is
∂²y/∂t² = -a ω² sin ωt = -ω² y.
Substituting into the wave equation gives
0 = (1/v²)(-ω² y) ⇒ ω² y = 0.
This can hold only if y ≡ 0 (trivial), ω = 0 or v → ∞, none of which describe a travelling wave. Hence y = a sin ωt does not satisfy the wave equation and does not represent a wave that propagates in space.

Q.2. A wave pulse is described by

Speed of wave

If the pulse has the form y = f(bx - ct) (or, more generally, depends on the combination bx ± ct), then the pulse shape is constant when the argument is constant. Setting bx - ct = constant gives
bx - ct = const ⇒ x = (c/b) t + const.
Thus dx/dt = c/b and the speed of the pulse is c/b. The sign of c/b (positive or negative) indicates direction according to the sign used in the argument.

Q.3. The displacement of a wave disturbance propagating in the positive x-direction is given by

and

where, x and y are in metre. The shape of the wave disturbance does not change during the propagation. What is the velocity of the wave?
Sol.

At t = 0, y is maximum at x = 0.
At t = 2 s, y is maximum at x = 1 m.
Hence the crest (or the point of maximum y) moves from x = 0 to x = 1 m in 2 s.
Distance travelled = 1 m, time = 2 s ⇒ v = (1 m)/(2 s) = 0.5 m/s in the positive x-direction.

Therefore, v = +0.5 m/s.

Q.4. A travelling wave pulse is given by,

Sol. When the argument of the pulse is of the form (x + vt) the pulse moves in the negative x-direction; when it is (x - vt) it moves in the positive x-direction. Here the coefficient of t and x have opposite signs in the argument, indicating motion towards negative x.
Speed of wave =

The amplitude is the maximum value of y. From the given expression the maximum value is 10/5 = 2 m.
Hence the pulse travels in the negative x-direction with speed shown by and amplitude = 2 m.

Q.5. If at f = 0, a travelling wave pulse on a string is described by the function,

Sol. For a pulse travelling along the positive x-direction with speed v, the argument containing x must be replaced by (x - vt) so that the shape translates in +x. Thus, if at t = 0 the pulse is y = f(x), then at time t it is y = f(x - vt).
Here v = 2 m/s, so replace x by (x - 2t) in the original expression.

Thus the coefficient of t in the argument becomes 2 times the coefficient of x (since v = 2 × 1 in SI units), giving the required wave function at time t.

Introductory Exercise 14.2

Q.1. The equation of a travelling wave is,
y(x, t ) 0.02 sin

(a) Identify the wave number k and angular frequency ω from the argument shown in

. The wave speed is v = ω/k. From the signs of kx and ωt (they have the same sign in the argument) the wave travels in the negative x-direction. Numerically, evaluating ω and k from the expression gives

Therefore v = -5 m/s (negative sign indicates motion in -x direction).
(b) Particle (transverse) velocity is ∂y/∂t. For y = A sin(kx - ωt) (or the given form),
v_p = ∂y/∂t = -A ω cos(kx - ωt).
Using the given values and substituting x = 0.2 m, t = 0.3 s we get

Numerically, v_p = A ω cos(θ) = 2 × cos(34) = 2 × (-0.85) = -1.7 m/s.
Thus the particle velocity at the specified point is -1.7 m/s (negative sign gives the direction).

Ques 2: Is there any relationship between wave speed and the maximum particle speed for a wave travelling on a string? If so, what is it?
Sol: As we know, Wave speed,

Q.3. Consider a sinusoidal travelling wave shown in the figure. The wave velocity is + 40 cm/s. Find :

From the figure, λ = 4 cm.

(a) Frequency f = v/λ = (40 cm/s)/(4 cm) = 10 Hz.
(b) Phase difference Δϕ between two points separated by Δx is Δϕ = (2π/λ) Δx. Thus
Δϕ = (2π/4 cm) × 2.5 cm = (π/2) × 2.5 = 1.25 π rad = 225°.
(c) Angular frequency ω = 2π f = 20π rad/s. A phase change of 60° = π/3 rad takes time t = (π/3)/ω = (π/3)/(20π) = 1/60 s ≈ 0.0167 s.
(d) The particle speed at any point is v_p = A ω cos(kx - ωt). At the mean position the particle speed is maximum in magnitude, v_p,max = A ω.
Using the values from the figure and the wave parameters gives

v = 125.7 cm/s = 1.26 m/s (magnitude). Considering the sign from the slope and wave direction, the particle velocity at P is v_p = -1.26 m/s.

Q.4. Transverse waves on a string have wave speed 12.0 m/s, amplitude 0.05 m and wavelength 0.4 m. The waves travel in the + x direction and at t = 0 the x = 0 ends of the string has zero displacement and is moving upwards.
(a) Write a wave function describing the wave.
(b) Find the transverse displacement of a point at x = 0.25 m at time t = 0.15 s.
(c) How much time must elapse from the instant in part (b) until the point at x = 0.25 m has zero displacement?
Sol.

Given v = 12.0 m/s, A = 0.05 m, λ = 0.4 m. Calculate wave number and angular frequency:
k = 2π/λ = 2π/0.4 = 5π rad/m.
f = v/λ = 12/0.4 = 30 Hz ⇒ ω = 2π f = 60π rad/s.
Since at t = 0 and x = 0 the displacement is zero and the point is moving upwards, the appropriate form is y = A sin(kx - ωt). Therefore

Thus the wave function is y(x,t) = 0.05 sin(5π x - 60π t) (in SI units).
(b) Put x = 0.25 m and t = 0.15 s into the wave function:
Argument = 5π(0.25) - 60π(0.15) = 1.25π - 9π = -7.75π.
sin(-7.75π) = sin(0.75π) = √2/2 ≈ 0.7071. Hence
y = 0.05 × 0.7071 = 0.0354 m = 3.54 cm.
(c) At the instant in part (b) the displacement is y = A/√2. The next time this point has zero displacement corresponds to a phase change Δϕ = π/4 (45°). The time required is Δt = Δϕ/ω = (π/4)/(60π) = 1/240 s ≈ 4.17 × 10^-3 s ≈ 4.2 ms.

Introductory Exercise 14.3

Q.1. Calculate the velocity of a transverse wave along a string of length 2 m and mass 0.06 kg under a tension of 500 N.
Sol.

 
Linear mass density μ = mass/length = 0.06 kg / 2 m = 0.03 kg/m.
Wave speed v = √(T/μ) = √(500 / 0.03) m/s = √(1.6667 × 10⁴) m/s ≈ 129.1 m/s.

Q.2. Calculate the speed of a transverse wave in a wire of 1.0 mm² cross-section under a tension of 0.98 N. Density of the material of wire is 9.8 × 10³ kg/m³.
Sol.

Cross-sectional area A = 1.0 mm² = 1.0 × 10^-6 m².
Linear mass density μ = ρ A = (9.8 × 10³ kg/m³) × (1.0 × 10^-6 m²) = 9.8 × 10^-3 kg/m.
Wave speed v = √(T/μ) = √(0.98 / 9.8 × 10^-3) = √(100) = 10 m/s.

Introductory Exercise 14.4

Q.1. Spherical waves are emitted from a 1.0 W source in an isotropic non-absorbing medium. What is the wave intensity 1.0 m from the source?
Sol.

For a point (spherical) source radiating power P uniformly into space, intensity at distance r is
I = P / (4π r²).
With P = 1.0 W and r = 1.0 m,
I = 1.0 / (4π × 1.0²) = 1 / (4π) W/m² ≈ 0.0796 W/m².

Q.2. A line source emits a cylindrical expanding wave. Assuming the medium absorbs no energy, find how the amplitude and intensity of the wave depend on the distance from the source?

Sol. For a line source the energy spreads cylindrically. At radius r from the line source the energy passes through a cylindrical surface of length L and circumference 2πr, so the area is proportional to r. Hence intensity I (power per unit area) falls as 1/r:
I ∝ 1/r.
For small-amplitude waves in a linear medium intensity is proportional to the square of the amplitude, I ∝ A². Combining these gives
A² ∝ 1/r ⇒ A ∝ 1/√r.
Therefore, for a cylindrical wave from a line source in a non-absorbing medium, intensity falls as 1/r and amplitude falls as 1/√r.